Problem: Simplify and expand the following expression: $ \dfrac{2n - 7}{5n - 10}-\dfrac{n - 1}{n + 3} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5n - 10)(n + 3)$ Multiply the first term by $\dfrac{n + 3}{n + 3}$ $ \begin{align*} \dfrac{2n - 7}{5n - 10} \times \dfrac{n + 3}{n + 3} & = \dfrac{(2n - 7)(n + 3)}{(5n - 10)(n + 3)} \\ & = \dfrac{2n^2 - n - 21}{(5n - 10)(n + 3)}\end{align*} $ Multiply the second term by $\dfrac{5n - 10}{5n - 10}$ $ \begin{align*} \dfrac{n - 1}{n + 3} \times \dfrac{5n - 10}{5n - 10} & = \dfrac{(n - 1)(5n - 10)}{(n + 3)(5n - 10)} \\ & = \dfrac{5n^2 - 15n + 10}{(n + 3)(5n - 10)}\end{align*} $ Now we have: $ = \dfrac{2n^2 - n - 21}{(5n - 10)(n + 3)} - \dfrac{5n^2 - 15n + 10}{(n + 3)(5n - 10)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2n^2 - n - 21 - (5n^2 - 15n + 10)}{(5n - 10)(n + 3)} $ $ = \dfrac{2n^2 - n - 21 - 5n^2 + 15n - 10}{(5n - 10)(n + 3)} $ $ = \dfrac{-3n^2 + 14n - 31}{(5n - 10)(n + 3)}$ Expand the denominator: $ = \dfrac{-3n^2 + 14n - 31}{5n^2 + 5n - 30}$